Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | + | <asy> | |
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
import graph; size(7cm); | import graph; size(7cm); | ||
Line 26: | Line 26: | ||
/* dots and labels */ | /* dots and labels */ | ||
dot((-1.1,4.92),dotstyle); | dot((-1.1,4.92),dotstyle); | ||
− | label(" | + | label("$A$", (-1.02,5.12), NE * labelscalefactor); |
dot((0.2,4.92),dotstyle); | dot((0.2,4.92),dotstyle); | ||
− | label(" | + | label("$B$", (0.28,5.12), NE * labelscalefactor); |
dot((-1.1,1.58),dotstyle); | dot((-1.1,1.58),dotstyle); | ||
− | label(" | + | label("$D$", (-1.02,1.78), NE * labelscalefactor); |
dot((1.04,1.58),dotstyle); | dot((1.04,1.58),dotstyle); | ||
− | label(" | + | label("$C$", (1.12,1.78), NE * labelscalefactor); |
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
/* end of picture */ | /* end of picture */ | ||
− | + | </asy> | |
If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. | If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. |
Revision as of 18:08, 1 December 2015
Problem
In trapezoid , and are perpendicular to , with , , and . What is ?
Solution
If and ,then . By the Pythagorean theorem, we have Solving the equation, we get .
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.